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vfiekz [6]
3 years ago
12

A rectangular current loop is located near a long straight wire that carries a current of 12A. The current in the loop is 5A. De

termine the magnitude of the net magnetic force that acts on the loop...
Physics
1 answer:
masya89 [10]3 years ago
8 0

Explanation:

The net magnetic force acting on the loop is the superposition of the forces on each side of the loop. The forces on the two sides perpendicular to the long straight wire will cancel (right hand rule). The magnitude of the magnetic force between two wires with currents I_1 = 12 A and I_2 = 5 A of length L = 0.5 m ( assuming not given in question).

F = \frac{\mu_o I_1 I_2 L}{2 \pi d}

so the forces on the near and far wires are

F_n = 2.7\times 10^{-4} N

F_f = 1.2\times 10^{-4} N

The force F_n is an attractive one, while F_f is a repulsive one (right hand rule). The magnitude of the force is therefore

F = F_n - F_f

F=1.5\times10^{-4}

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A car is traveling at 39.7 mi/h on a horizontal highway. The acceleration of gravity is 9.8 m/s 2 . If the coefficient of fricti
777dan777 [17]

Answer:

The minimum distance in which the car will stop is

x=167.38m

Explanation:

39.7\frac{mi}{h}*\frac{1km}{0.621371mi}*\frac{1000m}{1km}*\frac{1h}{3600s}=17.747\frac{m}{s}

∑F=m*a

∑F=u*m*g

The force of friction is the same value but in different direction of the force moving the car so it can stop so

F=m*a\\a=\frac{F}{m}\\a=\frac{u*m*g}{m}\\a=u*g\\a=0.096*-9.8\frac{m}{s^{2} }

a=-0.9408 \frac{m}{s^{2}}

v_{f}^{2}=v_{o}^{2}+2*a*(x_{f}-x_{o})\\v_{f}=0 \\x_{o}=0\\0=v_{o}^{2}+2*a*x_{f}\\x_{f}=\frac{v_{o}^{2}}{2*a} \\x_{f}=\frac{(-17.747\frac{m}{s})^{2}}{2*(-0.9408)} \\x_{f}=167.38m

4 0
3 years ago
A man does 4,475 J of work in the process of pushing his 2.50 103 kg truck from rest to a speed of v, over a distance of 26.0 m.
Tcecarenko [31]

Answer:

a) 1.89 m/s  b) 172.1 N

Explanation:

a)

  • Applying the work-energy theorem, if we can neglect the friction between truck and road, the total change in kinetic energy must be equal to the work done by the external forces.
  • This work, is just 4,475 J.
  • So we can write the following equation:

        \Delta K = \frac{1}{2} * m*v^{2} = 4,475 J

  • where m= mass of the truck = 2.5*10³ kg.
  • So, we can find the speed v, as follows:

        v =\sqrt{\frac{2*W}{m}} =\sqrt{\frac{2*4,475J}{2.5e3kg} }  = 1.89 m/s

b)

  • The work done by the man, is just the horizontal force applied, times the displacement produced by the force horizontally:

        W = F*d

  • We can solve for F, as follows:

        F = \frac{W}{d} = \frac{4,475 J}{26.0m} =  172.1 N

4 0
3 years ago
What is the difference between AM radio waves and FM radio waves?
N76 [4]
The correct answer is B.
6 0
3 years ago
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A bullet is shot from a rifle with a speed of 720 m/s. What time is required for the bullet to strike a target 3240 meters away
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4.5 seconds you devoured 3240/720=4.5
7 0
3 years ago
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Solve for y<br>11=y/3+7​
butalik [34]

Answer:

Explanation:

11 = y/3+7

collecting like terms

11 - 7 = y/3

4 = y/3

multiplying both sides by 3 gives us

12 = y

therefore

y = 12

3 0
3 years ago
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